Rewriting $\sin(\arccos(y))$ and $\arcsin(\cos(x))$ Prove the following identities:
\begin{align}
(a) && \sin(\arccos(y)) &= \sqrt{1-y^2}\\
(b) && \arcsin(\cos(x)) &= \frac{\pi}{2}-x
\end{align}
For (a) I am not sure how I would get a root out of any identity.
For (b) I can transform the $\cos$ to $\sin(x + \pi / 2)$ so I would assume the result
will be $x + \pi / 2$, and not $\pi / 2 - x$.
 A: For 'a':
$$\sin\left(\arccos(y)\right)=\sqrt{1-\cos^2\left(\arccos(y)\right)}=$$
$$\sqrt{1-\cos\left(\arccos(y)\right)\cos\left(\arccos(y)\right)}=\sqrt{1-yy}=\sqrt{1-y^2}$$
HINT (for 'b'):
$$\arcsin(\cos(x))\ne\frac{\pi}{2}-x$$

A: Hint:
$\arccos(y)=x$ means that $\cos x= y$
so $\sin (\arccos (y))=\sin x$ and $\sin x=\sqrt{1-\cos^2 x}$
You can do the same for the other identity:
$\arcsin(\cos x)=y \iff \sin y=\cos x$, than
 use $\sin(\pi/2 -x)=\cos x$.
A: This answer pertains to part (b).
If it were true that $\arcsin(\sin\theta)=\theta$ for all $\theta$, then the fact that $\cos x=\sin(x+\pi/2)$ would imply $\arcsin(\cos x)=\arcsin(\sin(x+\pi/2))=x+\pi/2$, as the OP observed.  But $\arcsin(\sin\theta)$ does not equal $\theta$ for all $\theta$.  Instead we have
$$\arcsin(\sin\theta)=\theta\quad\text{if and only if }-\pi/2\le\theta\le\pi/2$$
What's actually going on here is that identity (b) has been stated incompletely.  It should say
$$\arcsin(\cos x)=\pi/2-x\quad\text{for }0\le x\le\pi$$
Note that if we let $\theta=\pi/2-x$, then 
$$0\le x\le\pi\implies\pi/2\le\theta\le\pi/2$$ and 
$$\sin\theta=\sin(\pi/2-x)=\cos x$$
Putting these together, we have
$$\arcsin(\cos x)=\arcsin(\sin\theta)=\theta=\pi/2-x$$
but only for $0\le x\le\pi$.
A: The value of $\arccos y$ is in the interval $[0,\pi]$, where the sine is positive. So, if $\alpha=\arccos y$, we know that $\cos\alpha=y$ and
$$
\sin\arccos y=\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-y^2}
$$
Similarly, $\arcsin$ has values in the interval $[-\pi/2,\pi/2]$. If $\alpha=\arcsin\cos x$, then $-\pi/2\le\pi/2$ and
$$
\sin\alpha=\cos x
$$
so
$$
\cos\left(\frac{\pi}{2}-\alpha\right)=\cos x
$$
Note that
$$
0\le\frac{\pi}{2}-\alpha\le\pi
$$
Thus, if $x\in[0,\pi]$, we can conclude that
$$
\arcsin\cos x=\frac{\pi}{2}-x
$$
Otherwise the relation is not true. For instance, if $x=\frac{3\pi}{2}$, we have $\cos x=0$ and so
$$
\arcsin\cos\frac{3\pi}{2}=\arcsin0=0\ne\frac{\pi}{2}-\frac{3\pi}{2}=-\pi
$$
